A body of mass 2 kg collides with a massless spring of force constant K = 4 N/m. The spring compresses by 1m. If coefficient of friction between the body and the surface is 0.1, the speed of the body at the time of collision is :
1. 2 m/s
2. 4 m/s
3. 6 m/s
4. 0.5 m/s
View Answer
The Kinetic Energy of the object will go to increase potential energy of the spring and a part of energy will be lost to friction
U1= ½ K x²= ½ × 4× 1² = 2J
Work done against friction = ü.mg.x= 0.1× 20 × 1= 2 J
So total kinetic energy is 4 = ½× 2 ×v ² ⇒ v = 2 m/s
The only force acting on a \(2\text{ kg}\) body as it moves along the positive x axis has component \(F_x = -6x\text{ N}\), where x is in metre. The velocity of the body at \(x = 3\text{ m}\) is \(8\text{ m/s}\). The velocity of the body at \(x = 4\text{ m}\) is:
1. \(9.2\text{ m/s}\)
2. \(5.7\text{ m/s}\)
3. \(7.4\text{ m/s}\)
4. \(6.6\text{ m/s}\)
View Answer
Applying the work-energy theorem, \(W = \Delta K ⇒ \int_{3}^{4} -6x , dx = \frac{1}{2} m(v_f^2 - v_i^2)\). Integrating gives \(-3(16 - 9) = \frac{1}{2} (2) (v_f^2 - 64)⇒ -21 = v_f^2 - 64\), which yields \(v_f = \sqrt{43} \approx 6.6\text{ m/s}\).
Given below are two statements:
Assertion (A): A block of mass \(m\) is dropped from a height \(h\) on to a spring of spring constant \(k\). The maximum compression \(x_{\text{max}}\) in the spring is given by \(x_{\text{max}} = \sqrt{\frac{2mgh}{k}}\)
Reason (R): The work done by gravitational force is equal to the elastic potential energy stored in the spring at maximum compression.
In the light of the above statements, choose the correct answer from the options given below.
1. Both (A) and (R) are true and (R) is the correct explanation of (A)
2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
3. (A) is true but (R) is false
4. (A) is false but (R) is true
View Answer
By conservation of energy, the total loss in gravitational potential energy is \(mg(h + x_{\text{max}})\), which equals the elastic potential energy stored in the spring \(\frac{1}{2}kx_{\text{max}}^2\). Thus, the given formula for \(x_{\text{max}}\) is incorrect as it neglects the term \(mgx_{\text{max}}\). Hence, (A) is false but (R) is true.
